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- Calculus proof of ln (ab)= lna - Mathematics Stack Exchange
My calculus book states the following theorem of the properties of natural logarithms: If a, b > 0 , then ln(ab)= lna + lnb The author goes on to prove this theorem as follows I do not understand
- natural logarithm of a complex number $\ln (a+bi)$
natural logarithm of a complex number $\ln (a+bi)$ Ask Question Asked 5 years, 5 months ago Modified 5 years, 5 months ago
- How is log base 10 related to the Natural Logarithm
This can similarly be expressed as ln7 divided by ln4, using the natural logarithm Using the above example, Why is it that these two ratios are equal, but log base 10 (7) is not equal to ln 7? More generally, why can the log base 10 (x) be replaced in the change of base calculation with lnx?
- How to show that $n^ {\ln (\ln (n))} = \ln (n)^ {\ln (n)}$
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- calculus - Power series representation of $\ln (1+x)$? - Mathematics . . .
I am reading an example in which the author is finding the power series representation of $\ln (1+x)$ Here is the parts related to the question: I think that I get everything except for one thing:
- Show that $\ln (a+b) =\ ln (a) + \ln (b)$ when $a = \frac {b} {b-1}$
The original question in my textbook asked if the equality $\ln (a+b) = \ln (a) + \ln (b)$ was true My initial answer was that it was not true, but I missed this special case, which I think was very hard to "see"
- Integration of $\ln\sin x$ from 0 to $ \frac {\pi} {2}$ by DUIS
Note that $\int\limits_0^ {\frac {\pi} {2}}\ln (\sin x)dx$ is an improper integral Without proving or at least assuming its existence it makes no sense to apply integration by substitution
- Why is the derivative of Ln (x) equal to 1 x with no domain restricted?
When we get the antiderivative of 1 x we put a absolute value for Ln|x| to change the domain so the domains are equal to each other But my question is then why do we not do this for the derivative
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